Math Problem Statement

Suppose that we have a sample space S = {E1, E2, E3, E4, E5, E6, E7}, where probabilities for each event and specific sets A, B, and C are provided. Tasks involve calculating probabilities of individual events, unions, and intersections, as well as determining whether events are mutually exclusive.

Solution

Let's go through the problem step by step. The image outlines the following tasks:

We are given a sample space S={E1,E2,E3,E4,E5,E6,E7}S = \{E_1, E_2, E_3, E_4, E_5, E_6, E_7\} with associated probabilities for each event:

P(E1)=0.10,P(E2)=0.20,P(E3)=0.15,P(E4)=0.25,P(E5)=0.10,P(E6)=0.05,P(E7)=0.15P(E_1) = 0.10, \quad P(E_2) = 0.20, \quad P(E_3) = 0.15, \quad P(E_4) = 0.25, \quad P(E_5) = 0.10, \quad P(E_6) = 0.05, \quad P(E_7) = 0.15

You are also given three events:

  • A={E1,E4,E5}A = \{E_1, E_4, E_5\}
  • B={E3,E6,E7}B = \{E_3, E_6, E_7\}
  • C={E4,E5,E7}C = \{E_4, E_5, E_7\}

Task (a): Find P(A)P(A), P(B)P(B), and P(C)P(C).

P(A)P(A)

P(A)=P(E1)+P(E4)+P(E5)=0.10+0.25+0.10=0.45P(A) = P(E_1) + P(E_4) + P(E_5) = 0.10 + 0.25 + 0.10 = 0.45

P(B)P(B)

P(B)=P(E3)+P(E6)+P(E7)=0.15+0.05+0.15=0.35P(B) = P(E_3) + P(E_6) + P(E_7) = 0.15 + 0.05 + 0.15 = 0.35

P(C)P(C)

P(C)=P(E4)+P(E5)+P(E7)=0.25+0.10+0.15=0.50P(C) = P(E_4) + P(E_5) + P(E_7) = 0.25 + 0.10 + 0.15 = 0.50

Task (b): What is ABA \cup B?

The union of AA and BB is the set of all elements that are in either AA or BB:

AB={E1,E3,E4,E5,E6,E7}A \cup B = \{E_1, E_3, E_4, E_5, E_6, E_7\}

Hence, P(AB)P(A \cup B) is:

P(AB)=P(E1)+P(E3)+P(E4)+P(E5)+P(E6)+P(E7)=0.10+0.15+0.25+0.10+0.05+0.15=0.80P(A \cup B) = P(E_1) + P(E_3) + P(E_4) + P(E_5) + P(E_6) + P(E_7) = 0.10 + 0.15 + 0.25 + 0.10 + 0.05 + 0.15 = 0.80

Task (c): What is ABA \cap B?

The intersection of AA and BB is the set of elements that are in both AA and BB:

AB=A \cap B = \varnothing

Since the intersection is the empty set, P(AB)=0P(A \cap B) = 0.

Task (d): Are events AA and CC mutually exclusive?

Two events are mutually exclusive if they have no elements in common, i.e., their intersection is the empty set. Let's find ACA \cap C:

AC={E4,E5}A \cap C = \{E_4, E_5\}

Since ACA \cap C \neq \varnothing, the events AA and CC are not mutually exclusive.


If you would like further details or clarification on any of the steps, feel free to ask! Here are 5 follow-up questions related to this problem:

  1. What is the probability of the complement of event AA?
  2. How do you calculate P(AC)P(A \cup C)?
  3. Are events BB and CC mutually exclusive?
  4. Can we apply the addition rule for P(AB)P(A \cup B)?
  5. How would you approach finding P(AC)P(A \cap C) if more events were added to AA or CC?

Tip: Remember, when calculating the union of two events, make sure not to double-count any elements that belong to both events!

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Math Problem Analysis

Mathematical Concepts

Probability Theory
Set Theory

Formulas

P(A) = P(E1) + P(E4) + P(E5)
P(B) = P(E3) + P(E6) + P(E7)
P(C) = P(E4) + P(E5) + P(E7)
P(A ∪ B) = P(E1) + P(E3) + P(E4) + P(E5) + P(E6) + P(E7)
P(A ∩ B) = 0 (empty set)

Theorems

Addition Rule of Probability
Mutual Exclusivity of Events

Suitable Grade Level

College level or advanced high school probability